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3 years ago

It is rate for any motion to

Physics

1 answer:

Gnoma [55]3 years ago

3 0

Answer:

a. stay the same for very long

Explanation:

It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

For motion to remain constant, there must not be a net force acting on the body
All the forces on the body must be balanced.
This is very hard to come by.
Motion changes very frequently.

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Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'

Ne4ueva [31]

Explanation:

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4 years ago

A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the centripetal acceleration of the truck?

asambeis [7]

Answer:

B. 25 m/s/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius of curvature.

a = v² / r

Given v = 10 m/s and r = 4 m:

a = (10 m/s)² / 4 m

a = 25 m/s²

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3 years ago

A cart, which has a mass of m - 2.50 kg, is sitting at the top of an inclined plane which is 3.30 meters long and which meets th

SashulF [63]

Answer:

Part(i) the time taken for this cart to reach the bottom of the inclined plane is 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane is 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane is 25.663 J

Explanation:

Given;

mass of the cart = 2.5 kg

angle of inclination, β = 18.5⁰

length of inclined plane = 3.3m

Part(i) the time taken for this cart to reach the bottom of the inclined plane

s = ut + ¹/₂×at²

initial vertical velocity, u = 0

s = 3.3 m

s = ¹/₂×at²

$t = \sqrt{\frac{2s}{a} }$

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

$t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s$

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane

V = a×t

V = 3.11 × 1.457 = 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane

KE = ¹/₂MV²

= ¹/₂ × 2.5× (4.531)²

= 25.663 J

6 0

4 years ago

Which nucleus completes the following equation?

Archy [21]

Answer:

I think B but I could be wrong

Explanation:

3 0

3 years ago

A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

Marrrta [24]

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003 m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

8 0

3 years ago